what is the meaning of this proof of this Lemma? 
There are two things I am confused about
1) In the second line of the proof, How does $a=0$ imply that $G$ contains the  $\epsilon-$dense set?
2) In the last line of the proof, How does the definition of $a$ imply that $r=0$?  
 A: Both questions are answered by the definition of $a$ as it happens!
For question 1, if $a = 0$ then because we define $a$ as the infimum of the set $\{t\in G | t >0\}$ we must have that there are elements of $G$ arbitrarily close to 0 (as Magdiragdag points out, not every positive $t$ need be in the group), and we can write $t$ as $\epsilon$ to make clear that we're talking about very small values of $t$.  All integer multiples of an element of $G$ are contained in the group, so $\mathbb{Z}\epsilon$ is contained in the group.  This means we have a countable (since $\mathbb{Z}$ is countable) set of elements of $G$ arbitrarily close to any other element of $G$, i.e. a dense subset of $G$.  Then $G$ being closed yields the equality with $\mathbb{R}$.
For question 2, by writing $r=s-pa$ we see that $r\in G$ (because $s$ and $pa$ are both elements of $G$), and we have stated that $0\leq r<a$.  However, since $a$ is the infimum of non-zero elements of $G$ by definition the only element of $G$ smaller than $a$ is $0$.  So $r=0$.
A: 1) 
If $a=0$ then for every $\epsilon>0$ there is $x\in G\cap (0,\epsilon)$. $G$ is group, thereby $\mathbb{Z}x\subset G$, and $\mathbb{Z}x$ is $\epsilon$-dense by definition.
2) 
Assume, that $r>0$. But $r\in \{t\in G: t>0\}$, and $r<a=\inf\{t\in G: t>0\}$. Contradiction.
